The reversible chemical reaction A+BC+D has the following equilibrium constant:

Question

The reversible chemical reaction A+BC+D has the following equilibrium constant: Kc=[C][D][A][B]=1.5

Part A

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Answer :

Ka = [C][D] / [A][B] = 1.5 ---------- (1)

A) Initially, [A] = [B] = 2.00 M. let at equilibrium 'X' M of A and B react to give 'X' M C and D

Hence at equilibrium,

[A] = [B] = (2.00 - X) M and [C] = [D] = 'X' M

Using these equilibrium concentration in eq.(1) we get,

(2.00-X)(2.00-X) / (X)(X) = 1.5

(2.00-X)2/X2 = 1.5

Taking square root of both sides,

(2.00-X) / X = 1.22

2.00 - X = 1.22X

2.00 = 1.22X + X

2.00 = 2.22X

X = 2.00 / 2.22

X = 0.90 M

Hence we have equilibrium concentrations as,

[A] = [B] = 2.00 - X = 2.00 - 0.90 = 1.10 M

[C] = [D] = 0.90 M

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Part B) Initially,

[A] = 1.00 and [B] = 2.00 M let at equilibrium 'Y' M of each reactant A and B react to give 'Y' M of C and D.

Hence, equilibrium concentrations are,

[A] = (1.00 - Y) M and [B] = (2.00 - B) M

[C] = [D] = 'Y' M

Using these equilibrium concentrations in eq.(1)

(2.00-Y)(1.00-Y)/(Y)(Y) = 1.5

2.00 - 3.00Y + Y2 / Y2 = 3./2 (1.5 = 3/2)

On dividendo,

(2.00 - 3.00Y)/ Y2 = 1/2

On cross multiplication,

Y2 = 2(2.00-3.00Y)

Y2 + 6.00 = 4.00

Let us solve this quadratic equation by perfect square methode.

Add to both sides of above eq. 9 (which makes LHS perfect square.

Y2 + 6.00 + 9= 4.00 + 9.

Y2 + 6.00 + 9 = 13.00

(Y+3.00)2 = 13.00

On taking square root of both sides,

Y + 3.00 = + 3.61 (only +ve square root taken as -ve will give -ve Y value and concentration cannot be-ve)

Y = 3.61 - 3.0

Y = 0.61

Hence equilibrium concentrations are,

[A] = 1.00 - Y = 1.00 - 0.61 = 0.39 M

[B] = 2.00 - 0.61 = 1.39 M

[C] = [D] = 0.61 M

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